Question

Iron reacts with chlorine gas to form iron (III) chloride. What mass of iron (III) chloride is expected if 5.00 g of iron is combined with 9.00 g of chlorine?

Answer 1

To find the mass of iron (III) chloride produced, first identify the limiting reactant, which in this case is iron, and then use stoichiometry to calculate the mass of the product. The expected mass of iron (III) chloride is 14.52 g.

Step-by-step explanation:

To calculate the mass of iron (III) chloride formed, we need to follow a few steps:

1. Write the balanced chemical equation for the reaction.
2. Determine the limiting reactant.
3. Use stoichiometry to find the mass of the product formed from the limiting reactant.

The balanced chemical equation is:

2 Fe (s) + 3 Cl₂ (g) → 2 FeCl₃ (s)

Next, we need to find the moles of each reactant. The molar mass of iron (Fe) is approximately 55.85 g/mol and that of chlorine (Cl₂) is approximately 70.90 g/mol. Therefore:

• 5.00 g Fe × (1 mol Fe / 55.85 g Fe) = 0.0895 mol Fe
• 9.00 g Cl₂ × (1 mol Cl₂ / 70.90 g Cl₂) = 0.1269 mol Cl₂

From the balanced equation, 2 moles of Fe react with 3 moles of Cl₂, thus:

• (0.0895 mol Fe) × (3 mol Cl₂ / 2 moles Fe) = 0.134 mol Cl₂ needed

Since we have more moles of Cl₂ available than needed, Fe is the limiting reactant. Now, we use the stoichiometry to find the mass of FeCl₃ produced:

• 0.0895 mol Fe × (1 mol FeCl₃ / 1 mol Fe) × (162.20 g FeCl₃ / 1 mol FeCl₃)
• = 14.52 g FeCl₃

Therefore, the expected mass of iron (III) chloride formed from 5.00 g of iron and 9.00 g of chlorine is 14.52 g.